Question: You have found the following ages (in years) of 5 turtles. Those turtles were randomly selected from the 46 turtles at your local zoo: $ 51,\enspace 30,\enspace 37,\enspace 88,\enspace 63$ Based on your sample, what is the average age of the turtles? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 46 turtles, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{51 + 30 + 37 + 88 + 63}{{5}} = {53.8\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {7.84} + {566.44} + {282.24} + {1169.64} + {84.64}} {{5 - 1}} $ {s^2} = \dfrac{{2110.8}}{{4}} = {527.7\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{527.7\text{ years}^2}} = {23\text{ years}} $ We can estimate that the average turtle at the zoo is 53.8 years old. There is also a standard deviation of 23 years.